Why is F[x] a UFD?

Question

When reading the proof for if $R$ is a UFD, then $R[x]$ is a UFD, the author uses a fact that $F[x]$ is a UFD. I don't quite understand this.

Why $F[x]$ is a UFD? ($F$ is the fraction field of $R$).

Can someone explain to me why? Thanks.

Answer 1

In the book Ideals, Varieties, and Algorithms by Cox D., Little J., O'Shea D. you can see :

Note that every PID is UFD.

Answer 2

We want to define a degree function for $F[X]$ and deduce that $F[X]$ is an euclidean domain and so an UFD.

Hou to construct our function?

Look at $\mathbb{Q}[x]$:

The constant terms have degree zero and the polynomials with leading monomial $x^n$ have degree $n$.

Let's set now the same values of degree function for $\mathbb{F}[x]$.

We have to prove:

• If $a$ and $b$ are in $\mathbb{F}[x]$ and $b$ is nonzero, then there are $q$ and $r$ in $\mathbb{F}[x]$ such that $a = bq + r$ and either $r = 0$ or $deg(r) < deg(b)$.
• For all nonzero $a$ and $b$ in $\mathbb{F}[x]$, we have $deg(a) \leq deg(ab)$

but now the proofs of these facts are the same as the case $\mathbb{Q}[x]$.